A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q0,V0,E0 and U0 respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one as

Question

A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q0,V0,E0 and U0 respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one as (A) Q>Q0  (B) V>V0  (C) E>E0  (D) U>U0

Tardigrade Admin · Accepted Answer

When dielectric slab is introduced capacity gets increased
while potential difference remains unchanged.

∴ V = V_{0}, C > C_{0}

Q = C V ∴ Q > Q_{0}

U = \frac{1}{2} C V^{2} ∴ U > U_{0}

E = \frac{V}{d}

but V and d both are unchanged
Therefore,

E = E_{0}

Therefore, correct options are (a) and (d).

$Q > Q_{0}$

$V > V_{0}$

$E > E_{0}$

$U > U_{0}$

Q>Q0​

V>V0​

E>E0​

U>U0​

When dielectric slab is introduced capacity gets increased while potential difference remains unchanged. ∴V=V0​,C>C0​ Q=CV∴Q>Q0​ U=21​CV2∴U>U0​ E=dV​ but V and d both are unchanged Therefore, E=E0​ Therefore, correct options are (a) and (d).

$Q > Q_{0}$

$V > V_{0}$

$E > E_{0}$

$U > U_{0}$